PHYS SCIENCES P1 GR12 MEMO SEPT2021 Xho English B Copy hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/IBANGA 12 SEPTEMBER 2021 PHYSICAL SCIENCES P1/INZULULWAZI P1 MARKING GUIDELINE/ISIKHOKELO SOKUMAKISHA MARKS/AMANQAKU 150 This marking guideline consists of 17 pages./ Esi sikhokelo sokumakisha sinamaphepha ali17.

Downloaded from hlayiso.com 2 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) GENERAL GUIDELINES/IMIQATHANGO EQHELEKILEYO 1. CALCULATIONS/UKUBALA 1.1 Marks will be awarded for: correct formula, correct substitution, correct answer with unit. Amanqaku ayakunikelwa ifomyula echanekileyo, ukufakela izimeli, iimpendulo ezichanekileyo ezineyunithi. 1.2 No marks will be awarded if an incorrect or inappropriate formula is used, even though there are many relevant symbols and applicable substitutions. Akukho manqaku azakunikwa ukuba kusetyenziswe ifomyula engeyiyo, nangona kukho iisimboli nezimeli ezizizo. 1.3 When an error is made during substitution into a correct formula, a mark will be awarded for the correct formula and for the correct substitutions, but no further marks will be given. Xa kwenzeke impazamo ekufakelweni kwezimeli kwifomyula echanekileyo, likhona inqaku elinganikelwa ifomyula nezimeli ezichanekileyo kodwa awakho amanye amanqaku anikwayo. 1.4 If no formula is given, but all substitutions are correct, a candidate will forfeit one mark. Ukuba akukho fomyula inikiweyo, kodwa zonke izimeli zichanekile, umlingwa uyakulahlekelwa linqaku. 1.5 No penalisation if zero substitutions are omitted in calculations where correct formula/principle is correctly given. Akuyi kuxhuzulwa manqaku ukuba izimeli zika zero zishiyelelwe kwizibalo apho ifomyula/umgaqo uchanekileyo. 1.6 Mathematical manipulations and change of subject of appropriate formulae carry no marks, but if a candidate starts off with the correct formula and then changes the subject of the formula incorrectly, marks will be awarded for the formula and correct substitutions. The mark for the incorrect numerical answer is forfeited. Uhlengahlengiso lwezibalo notshintsho lwesubject kwifomyula ezizizo alunamanqaku, kodwa ukuba umlingwa uqale ngefomyula echanekileyo ze watshintsha isubject yefomyula ngokuphosakeleyo, amanqaku ayakunikelwa ifomyla nezimeli ezichanekileyo. Inqaku lempendulo engachanekanga liyakumphosa. 1.7 Marks are only awarded for a formula if a calculation has been attempted, i.e. substitutions have been made or a numerical answer given. Amanqaku anikelwa ifomyula kuphela xa isibalo sizanyiwe, izimeli zifakiwe okanye impendulo elinani inikiwe. Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 3 1.8 Marks can only be allocated for substitutions when values are substituted into formulae and not when listed before a calculation starts. Amanqaku ayakunikezelwa kuphela ukuba amaxabiso afakiwe kwifomyula kwaye azidweliswanga nje phambi kokubala. 1.9 All calculations, when not specified in the question, must be done to a minimum of two decimal places. Zonke izibalo ukuba akuqononondiswanga kumbuzo, mazenziwe zibenamaqondo amabini edesimali . 1.10 If a final answer to a calculation is correct, full marks will not automatically be awarded. Markers will always ensure that the correct/appropriate formula is used and that workings, including substitutions, are correct. Ukuba impendulo yokugqibela ichanekile, amanqaku apheleleyo awakusuka anikelwe. Abamakishi bakuqinisekisa ukuba ifomyula echanekileyo isetyenzisiwe kwaye umsebenzi kunye nezimeli zichanekile. 1.11 Questions where a series of calculations have to be made (e.g. a circuit diagram question) do not necessarily always have to follow the same order. FULL MARKS will be awarded provided it is a valid solution to the problem. However, any calculation that will not bring the candidate closer to the answer than the original data, will not count any marks. Kwimibuzo apho kulindeleke uhlohlo lwezibalo (umz. Umbuzo wecircuit diagram) akunyanzelekanga lulandelelwano lunye. AMANQAKU APHELELEYO ayakunikwa ukuba sisisombululo esisiso kumbuzo. Nasiphi na isibalo esingamsondeziyo umlingwa kwimpendulo echanekileyo asiyi kunikwa manqaku. 2. UNITS/IIYUNITHI 2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question. Abalingwa bayakuhluthelwa amanqaku kube kanye ukuba bathe gqolo besebenzisa iyunithi engachanekanga kumbuzo lowo. 2.2 Units are only required in the final answer to a calculation. Iiyunithi zilindeleke kwimpendulo yokugqibela yesibalo. 2.3 Marks are only awarded for an answer, and not for a unit per se. Candidates will therefore forfeit the mark allocated for the answer in each of the following situations:  Correct answer + wrong unit  Wrong answer + correct unit  Correct answer + no unit Amanqaku anikezelwa impendulo, hayi iyunithi. Umlingwa ke ngoko uyakulahlekelwa ngamanqaku kwezi meko ziladelayo:  Impendulo echanekileyo + iyunithi ephosakeleyo  Impendulo engachanekanga + iyunithi echanekileyo  Impendulo echanekieyo + iyunithi eshiyelelweyo/engekhoyo Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 4 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) 2.4 SI units must be used except in certain cases, e.g. V .m-1 instead of N.C-1, and cm•s-1 or km•h-1 instead of m•s-1 where the question warrants this. SI-yunithi mazisetyenziswe, ngaphandle kwa kwezi meko,umz. V•m-1 endaweni ka N•C-1, kunye no cm•s-1 ka km•h-1 endaweni ka m•s-1 kodwa umbuzo uyakungqina oku. 3. GENERAL/OKUQHELEKILEYO 3.1 If one answer or calculation is required, but two are given by the candidate, only the first one will be marked, irrespective of which one is correct. If two answers are required, only the first two will be marked, etc. Ukuba kulindeleke impendulo okanye isibalo esinye ze umlingwa anike zibembini, kuyakuqwalaselwa kuphla le yokuqala nokuba yeyiphi kuzo echanekileyo. Ukuba kulindeleke iimpendulo ezimbini, kuyakuqwalaselwa kuphela isibini sokuqala. 3.2 For marking purposes, alternative symbols (s, u, t, etc.) will also be accepted. Ukwenzela ukumakisha, ezinye iisimboli (s, u, t, ens.) ziyakwamkeleka. 3.3 Separate compound units with a multiplication dot, not a full stop, for example, m•s-1. For marking purposes, m•s-1 and m/s will also be accepted. Yohlula iiyunithi ezidityanisiweyo ngemultiplication dot. Hayi isingxi, umz. m•s-1. Ukwenzela ukumakisha, u m•s-1 no m/s bayakwamkeleka nabo. 4. POSITIVE MARKING Positive marking regarding calculations will be followed in the following cases: Positive marking malunga nezibalo iyakulandelwa kwezi meko zilandelayo: 4.1 Sub-question to sub-question: When a certain variable is calculated in one sub-question (e.g. 3.1) and needs to be substituted in another (3.2 of 3.3), e.g. if the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3, full marks are to be awarded for the subsequent sub-questions. Umbuzwana ukuya kumbuzwana: xa ivariable ethile ibalwe kumbuzwana omnye (umz. 3.1) kwaye idinga ukufakelwa kwenye (3.2 okanye 3.3), umz. ukuba impendulo ka 3.1 ayichanekanga ibe ifakelwe kakuhle ku 3.2 okanye 3.3,amanqaku apheleleyo ayaunikelwa umbuzwana olandelayo. 4.2 A multistep question of a sub-question: If the candidate has to calculate, for example, current in die first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. Umbuzo onamanqanaba amaninzi wombuzwana: Uba umlingwa ufanele abale. umz. Umsinga kwibakala lokuqala aze aliphose ngenxa yezimeli ezingachanekanga, inqaku lesimeli nelempendulo yokugqibela ziyakumphosa. Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 5 5. NEGATIVE MARKING Normally an incorrect answer cannot be correctly motivated if based on a conceptual mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the answer given in QUESTION 3.1, and QUESTION 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for e.g. QUESTION 3.1 is based on a calculation, the motivation for the incorrect answer could be considered. Ngokwesiqhelo impendulo engachanekanga ayithetheleleki isekelezelwe kwingcamangao enempazamo. Ukuba umlingwa ulindeleke ukuba acacise/asekele kuMBUZO 3.2 impendulo abeyinike ku MBUZO 3.1, ube loMBUZO 3.1 ungachanekanga, akukho manqaku azakunikelwa uMBUZO 3.2. Ukuba impendulo ekuMBUZO 3.1 isekelezelwe ekubaleni, imotivation yempendulo engachanekanga ingajongwa. Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 6 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) QUESTION/UMBUZO 1: MULTIPLE-CHOICE QUESTIONS/ IMIBUZO EKHETHISAYO 1.1 C  (2) 1.2 A  (2) 1.3 B  (2) 1.4 D  (2) 1.5 D  (2) 1.6 C  (2) 1.7 A  (2) 1.8 C  (2) 1.9 D  (2) 1.10 B  (2) [20] Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 7 QUESTION/UMBUZO 2 2.1.1 When a (non-zero) resultant/net force acts on an object, it accelerates in the direction of the force. The acceleration is directly proportional to the force and inversely proportional to the mass of the object.  Xa i(non-zero) resultant/net force isebenza kwi object, ibalekela kwicala leforce. Isantya si directly proportional kwiforce kwaye si inversely proportional kubunzima be object  (2) 2.1.2 OPTION 1 OPTION 2 N N F Fy cc  f f T Fx  T w w (5) Accept the following symbols./yamkela ezi simboli zilandelayo: N FN/Normal/eqhelekileyo/Normal force/amandla aqhelekileyo f Ff / fk / fr / frictional force/ /kinetic frictional force / w Fg,/mg/weight/FEarth on block/49 N/gravitational force/ T Tension / FT Fapplied  F / Applied force Marks awarded for arrow and label/amanqaku anikelwa itolo negama. Do not penalise for length of arrows since drawing is not drawn to scale. Sukuhlutha manqaku ukuba itolo lifutshane/lide (umzobo awenziwanga ngokweskali). 3 Any other additional force(s) 4 3 If force(s) do not make contact with body. Max./Maks. 4 Ukuba iiforce aziyichananga/aziyithintanga iobject. Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 8 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) 2.1.3 OPTION 1 OPTION 2 (To the right is positive) (To the right is negative) (ukuya ekunene positive) (ukuya ekunene inegative) Fnet = ma Fnet = ma Any one  Any one T – f = ma f – T = -ma Enige een  Enige 1 Fcosθ - T – f = ma T + f – Fcosθ = -ma T – 10 = 2 (2) 10 – T = 2(-2)  T = 14 N T = 14 N Fcosθ - T – f = ma T + f – Fcosθ = -ma ° 14 + 15 -Fcos20 = 5(-2)  Fcos20o – 14 – 15 = 5(2)  F = 41,50 N  F = 41,50 N  (5) 2.2 Gm 1m 2 F=  d2 (6,67  10-11 )( 5,98  1024 )(200) 1 842,50  =  d2 d = 6 579 982,80 m distance above earth surface/umgama ngaphezu komhlaba = 6 579 982,80 – 6,38 x 106  = 199 982,80 m (1,9998280 x 105 m / 2,00 x 105 m)  (5) [17] QUESTION/UMBUZO 3 3.1.1 OPTION 1 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf2 = vi2 + 2aΔy  vf2 = vi2 + 2aΔy  (-25)2 = (-20)2 + 2(-9,8) Δy  (25)2 = (20)2 + 2(9,8) Δy  Δy = -11,48 Δy = 11,48 m  Δy = 11,48 m  OPTION 2 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt vf = vi + aΔt -25 = -20 + (-9,8)(Δt) 25 = 20 + (9,8)(Δt) Δt = 0,51 s Δt = 0,51 s Δy = viΔt + ½gΔt2  Δy = viΔt + ½gΔt2  Δy = (-20)(0,51) + ½(-9,8)(0,51)2  Δy = (20)(0,51) + ½(9,8)(0,51)2  Δy = -11,47 Δy = 11,47 m  Δy = 11,47 m  Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 9 OPTION 3 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt vf = vi + aΔt -25 = -20 + (-9,8)(Δt) 25 = 20 + (9,8)(Δt) Δt = 0,51 s Δt = 0,51 s v  vi v  vi Δy = f Δt  Δy = f Δt  2 2 - 25  (-20) 25  20 Δy = x 0,51 Δy = x 0,51 2 2 Δy = -11,48 m Δy = 11,48 m  Δy = 11,48 m  OPTION 4 (ACCEPT/YAMKELA) UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt vf = vi + aΔt -25 = -20 + (-9,8)(Δt) 25 = 20 + (-9,8)(Δt) Δt = 0,51 s Δt = 0,51 s Δy = lb + ½bh  Δy = lb + ½bh  Δy = 20 x 0,51 + ½(5)0,51  Δy = 20 x 0,51 + ½(5)0,51  Δy = 11,48 m  Δy = 11,48 m  (3) 3.1.2 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt  vf = vi + aΔt  -25 = -20 + (-9,8)(Δt)  25 = 20 + (-9,8)(Δt)  Δt = 0,51 s (time to reach to ground) Δt = 0,51 s (time to reach to ground) (ixesha ukuya kufika (ixesha ukuya kufika emhlabeni) emhlabeni) vf = vi + aΔt 0 = 12 + (-9,8)(Δt)  vf = vi + aΔt Δt = 1,22 s 0 = -12 + (9,8)(Δt)  (time to reach maximum height) Δt = 1,22 s (ixesha ukufika phezulu) (time to reach maximum height) t = 1,22 + 0,51  (ixesha ukufika phezulu) t = 1,73 s  t = 1,22 + 0,51  t = 1,73 s  (5) 3.1.3 OPTION 1 Positive marking from 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf2 = vi2 + 2aΔy  vf2 = vi2 + 2a  02 = 122 + 2(-9,8) Δy  02 = -122 + 2(9,8) Δy  Δy = 7,35 m Δy = -7,35 m Displacement = - 11,48 + 7,35  Displacement = 11,48 + (- 7,35)  = - 4,13 = 4,13 m = 4,13 m (downwards/ukuhla)  (downwards/ukuhla)  Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 10 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) OPTION 2 Positive marking from 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt vf = vi + aΔt 0 = 12 + (-9,8)(Δt) 0 = -12 + (9,8)(Δt) Δt = 1,22 s Δt = 1,22 s Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  Δy = 12 x1,22 + ½ (-9,8)(1,22)2  Δy = (-12)(1,22) + ½ (9,8)(1,22)2  Δy = 7,35 m Δy = -7,35 Displacement = - 11,48 + 7,35  Displacement = 11,48 + (- 7,35)  = - 4,13 = 4,13 m = 4,13 m (downwards/ukuhla)  (downwards/ukuhla)  OPTION 3 Positive marking from 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt vf = vi + aΔt 0 = 12 + (-9,8)(Δt) 0 = -12 + (9,8)(Δt) Δt = 1,22 s Δt = 1,22 s v  vi v  vi Δy = f Δt  Δy = f Δt  2 2 0  (12) 0  (-12) Δy = x 1,22  Δy = x 1,22  2 2 Δy = 7,32 m Δy = - 7,32 m Displacement = - 11,48 + 7,32  Displacement = 11,48 + (- 7,32)  = - 4,16 = 4,16 m = 4,16 m (downwards/ukuhla)  (downwards/ukuhla)  OPTION 4 Positive marking from 3.1.1 UPWARDS POSITIVE UPWARDS NEGATIVE UKUNYUKA POSITIVE UKUNYUKA NEGATIVE vf = vi + aΔt vf = vi + aΔt 0 = 12 + (-9,8)Δt 0 = -12 + (9,8)Δt Δt = 1,22 s Δt = 1,22 s Area = ½bh  Area = ½bh  = ½ (1,22)(12)  = ½ (1,22)(-12)  = 7,32 m = -7,32 m Displacement = - 11,48 + 7,32  Displacement = 11,48 + (- 7,32)  = - 4,16 = 4,16 m = 4,16 m (downwards/ukuhla)  (downwards/ukuhla)  (4) Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 11 3.2 Positive marking from 3.1.1 and 3.1.2 Position/Posisie(m) 11,48 m Time/Tyd(s) 1,73 m CRITERIA FOR MARKING/ Correct shape/ishape echanekileyo  Height indicated/umphakamo ubonisiwe  (11,48 m) Time t indicated/ixesha t libonisiwe (1,73 s)  (3) [15] QUESTION/UMBUZO 4 4.1 In an isolated system total linear momentum is conserved.  (2) 4.2.1 ∑pi = ∑pf Any one/nayiphi  mAviA + mBviB = (mA + mB)vf (2 x viA) + (4 x -5) = (2+4)(-1,67)  viA = 4,99 m.s-1 (East/Empuma)  (4) 4.2.2 POSITIVE MARKING FROM 4.2.1 POSITIVE MARKING FROM 4.2.1 OPTION 1 OPTION 2 Fnet.∆t = ∆p Fnet.∆t = ∆p Any one Any one Fnet.∆t = m(vf – vi) /nayiphi  Fnet.∆t = m(vf – vi) /nayiphi  Fnet x 0,01 = 2 (-1,67- 4,99)  Fnet x 0,01 = 2 [1,67- (-4,99)]  Fnet = - 1 332 N Fnet = - 1 332 N Fnet = 1 332 N west/left/entshona Fnet = 1 332 N west/entshona  kwesokunxele  Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 12 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) OPTION 3 OPTION 4 Fnet = ma Any one Fnet = ma Any one v f - vi /nayiphi  v - vi Fnet = m ( ) Fnet = m ( f ) / nayiphi  t t - 1,67 - 4,99  1,67 - (- 4,99)  Fnet = 2 x ( ) Fnet = 2 x ( ) 0,01  0,01 Fnet = -1 332 Fnet = 1 332 N west/ Fnet = 1 332 N entshona/kwesokunxele  west/left/entshona/kwesokunxele  OPTION 5/OPSIE 5 OPTION 6/OPSIE 5 Fnet.∆t = ∆p Fnet.∆t = ∆p Any one / Any one / Fnet.∆t = m(vf – vi) nayiphi Fnet.∆t = m(vf – vi) nayiphi Fnet(0,01)  = 4(-1,67 – -5)  Fnet(0,01)  = 4(1,67 – 5)  Fnet = 1 332 Fnet = -1 332 FAB = - FBA FAB = - FBA Fnet(BA) = 1 332 N west/left  Fnet(BA) = 1 332 N west/left  (4) [10] QUESTION/UMBUZO 5 5.1 Gravitational force  (1) 5.2 12 ∆x =  = 24 m sin30o Wf = f.∆x cos θ  Wf = 35,5 x 24 cos 180o  Wf = - 852 J  (4) 5.3 Zero/0 J  (1) Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 13 5.4 Positive marking from 5.2 OPTION 1 Wnet = Ek Any one/nayiphi  Wf + W F + W Fg = Ek f x ∆x cos θ + F xcos θ + mg(h2 – h1) = Ek -852  + (62,5 x 24 cos180o)  + m(9,8)(12-0)  = 0 m = 20 kg  OPTION 2 Wnc = Ep + Ek W f + W F = Ep + Ek Any one/ nayiphi f∆xcos θ + F xcos = mg(h2 – h1) + Ek entshona/kwesokunxele o -852 + (62,5)(24)cos180 = m(9,8)(0 – 12) + 0 m = 20 kg  OPTION 3 Wnet = Ek W f + W F + W w = Ek Any one/ nayiphi  f∆xcos θ + F xcos θ + mgΔxcos θ = Ek -852  + (62,5)(24)cos180o  + m(9,8)(24)cos60o  = 0 m = 20 kg  (5) [11] QUESTION/UMBUZO 6 6.1.1 520 Hz / 520 waves per second (waves.s-1)  (1) 6.1.2 The change in frequency (or pitch) observed/detected by a listener because the listener and the sound source have different velocities relative to the medium of sound propagation.  OR/OKANYE The (apparent) changed in observed/detected frequency (pitch) as a result of relative motion between the sound source and the listener.   (2) 6.1.3 TOWARDS  Detected frequency is higher than the source frequency  (2) 6.1.4 v ± vL fL = fs  v ± vs 343 520  =  (480)  343 - v s vs = 26,38 m·s-1  (5) Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 14 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) 6.1.5 Decreases/Iyehla For a constant velocity/speed of sound, if the frequency increases, λ decreases.  OR λα at constant velocity/speed  OR f α at constant velocity/speed  (2) 6.2 Light from the star is shifted towards longer wavelength (towards the red end of the spectrum)  (2) 6.3 Used to measure the direction and speed of blood flow in arteries and veins.  Isetyenziselwa ukubona apho igazi liya khona nesantya elihamba ngaso emithanjeni OR/OKANYE Used to measure the heartbeat of a foetus in the womb. isetyenziselwa ukujonga ukubetha kwentliziyo yosana olusesibelekweni (1) [15] QUESTION/UMBUZO 7 7.1.1 GAIN/IFUNYENWE  (1) 7.1.2 Q n=  qe 5 x 10 -6 n= 19  1,6 x 10 - n = 3,125 x 1013 (electrons)  (3) 7.1.3 kQ E=  r2 9 x 109 x 5 x 10 -5 E=  0,12  E = 4,5 x 106 N.C-1  left/ekunxelE  (5) 7.2.1 Negative  Like charges repel each other/Ezifanayo ziyakhabana.  OR/OKANYE The charges repel each other. If sphere A is negative, then sphere B must also be negative.  icharges ziyakhabana. Ukuba usphere A unegative, ngoko usphere B naye makabe negative. (2) Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 15 7.2.2 FE = Tsin30o FE = 25sin30o  FE = 12,5 N kQ 1Q 2 FE =  r2 (9  109 )( 5  10 -6 ) Q 12,5 =  0,05 2  Q = 6,94 x 10-7 C  (6) [17] QUESTION/UMBUZO 8 8.1.1 Temperature/ubushushu Length of the conductors/ubude becoductors  (Any two/nasiphi isibini) Thickness of the conductors/ukutyeba kweconductors. ACCEPT: Type of material YAMKELA: uhlobo lwematerial (2) 8.1.2 Gradient is the inverse of the resistance. / OR Gradient =  (1) 8.1.3 Conductor C./Geleier C.  It has the highest resistance. The higher the resistance of a conductor, the more heat is produced in the conductor if the current is constant.  (2) 8.2.1 R =  12 R=  1,5 R=8Ω (3) 8.2.2 OPTION 1 OPTION 2 Rtotal = R + r  = I(R+ r)  [8 = (4 + 3) + r ] 12 = 1,5 [(4 + 3) + r]  r=1Ω r=1Ω (4) 8.2.3 W = I2R∆t  W = (1,5)2(3)(180)  W = 1 215 J  (3) 8.3.1 Decrease/iyehla.  (1) 8.3.2 Increase/iyenyuka.  (1) Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com 16 PHYSICAL SCIENCES/INZULULWAZI P1 (EC/SEPTEMBER 2021) 8.4 Increase/iyenyuka.  Rext decreases. Current through battery increases.  W = I2r∆t / Energy transfer to the battery/work done by battery increases.  (3) [20] QUESTION/UMBUZO 9 9.1 Mechanical energy to electrical energy. (2) 9.2 AC generator has slip rings and DC generator has a split ring / commutator  (1) 9.3 Potential Difference (emf) (V) Potensiaalverskil (emk) CRITERIA FOR MARKING Correct shape  Axes labelled correct  Vmax indicated on graph/  (2) 9.4 Pave = VrmsIrms  Pgem = VwgkIwgk  2000 = Irms x 230  2000 = Iwgk x 230  Irms = 8,70 A Iwgk = 8,70 A I I Irms = max Iwgk = max 2 2 Imax I 8,70 =  8,70 = max  2 2 Imax = 12,30 A  Imaks = 12,30 A  (4) 9.5 Vmax Vrms =  2 Vmax 230 =  2 Vmax/Vmaks = 325,27 V  (3) [12] Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) PHYSICAL SCIENCES/INZULULWAZI P1 17 QUESTION/UMBUZO 10 10.1 Work function (of the metal))  hc Ek(max) = – W0   The intercept on the vertical axis = W 0. OR hc = W 0 + Ek(max)   The intercept on the vertical axis is equal to the W o (3) 10.2 E = W 0 + Ek(max) hf = W 0 + Ek(max) Any one/nayiphi  hf = hf0 + Ek(max) 6,63 x 10-34 x 6,16 x 1014 = 6,63 x 10-34 f0  + 5,6 x 10-20  f0 = 5,32 x 1014 Hz  (5) 10.3.1 Remain the same/ayitshintshi  The gradient is equal to the product of Planck’s constant and the speed of light in vacuum which are constants.  OR Gradient = hc, which are constants (2) 10.3.2 Remains the same/Ayitshintshi  Ek(max) (J) Intensity CRITERIA FOR MARKING Axes labelled  Correct shape  (3) [13] TOTAL/EWONKE: 150 Copyright reserved/Akuvumelekanga ukufotokopa Please turn over/Tyhila